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Dave the bass
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#31

Post by Dave the bass »

Hello, me again.

I just can't stop multiplying and dividing :) but get this...

http://www.r-type.org/pdfs/1626.pdf

The RCA data sheet for a 1626, I've actually got 4 of these and thought I'd try and exercise my new found powers of mathematics buuuuuuut I can see the mu (amplification factor) but can't see either the gm or the anode (plate) resistance.

Where do I start? other chaps have used this val;ve very succesfully in ccts so I'm guessing they must've worked it all out (surely)!

Puzzled of Dartford.

DTB
"The fat bourgeois and his doppelganger"
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Nick
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#32

Post by Nick »

Ok,, all the information you need is on the anode curves. Pick a op point (I will use the 250v 25ma on in the spec, and draw a tangent to the grid curve at that point.

Then the slope of the line is the ra of the valve at that point. So the slop is 310v and 120ma, so using ohms law again ra = 310v / .12A = 2580 ohm
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#33

Post by Andrew »

I started typing "its all in the curves" but Nick's a much better teacher!

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andrew Ivimey
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#34

Post by andrew Ivimey »

I'm not the devil's advocate here but I don't often 'do the maths' even when I can, but as you mention the 1626, well I just went along with the Darling idea and found that 5K output trafo was just the job (for a huge 0.8watt or so output; flippin'eck eh) and the interstage had windings of 10K, the driver valve was also a 1626, though by now we all know, there are many and varied suitable drivers, all worth a pot.

Bin go.

I'm very much enjoying your questions Dave and of course Nick's answers are spot on - so I'm filling in lots of gaps as you go along.

onwards!
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#35

Post by simon »

andrew Ivimey wrote:I'm very much enjoying your questions Dave and of course Nick's answers are spot on - so I'm filling in lots of gaps as you go along.
I'll second that Andrew, and bravo Dave, you're on the way :).
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Dave the bass
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#36

Post by Dave the bass »

Sorry Nick, I'm totally lost :-( and I know you've gone through this in the past with me and others but this bit just isn't going in. Sorry about this.

OK, I can see on the 1626 data sheet that they state max anode voltage is 250V and max anode current is 25mA. I can also see on your gif where that point on the curve on the -32 grid voltage line is but I'm at a loss how you've got that other point, the one thats sitting on the 0mA line at approx 185V. I know its summat to do with ohms law and you've gone through it before but I haven't got it yet.

I thought I'd cracked it for minute and could just use the ohms law triangle 250V/0.025A to give me the ra but it works out as 10K Ohm :hmph:

Dissillusioned of Dartford.
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Dave the bass
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#37

Post by Dave the bass »

simon wrote:
andrew Ivimey wrote:I'm very much enjoying your questions Dave and of course Nick's answers are spot on - so I'm filling in lots of gaps as you go along.
I'll second that Andrew, and bravo Dave, you're on the way :).
Ta fellas, as you've problee worked out its an uphill struggle for me. Breaking it down into number triangles helps me over come the problems i have with maths but the daft thing right now is that I can't see where the numbers are actually coming from on a graph sitting right here in front of me.

I'm beginning to think thats half my problem, its not so much the adding/subtraction/divide and multiply (as long as I've got access to a calculator) but where the numbers ACTUALLY come from in the first place.

Today I've proved to myself I can multiply and divide on a calculator again, i got this far on another thread (2A3 amp build i think) and then fell over on the number work when it came to the curves but today was a giant leap for bass-kind, the gm triangle is ok now! wanna know an ra? I'm yer fella :)

Its just these curves and extracting numbers from them that is stuffing me at the moment.

DTB
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#38

Post by al newall »

Don't worry Dave, i couldn't work it out either.
By the way you should have a PM.
Much to learn there is.
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#39

Post by Nick »

OK, all I did, and I guess there was a couple of steps in one was try and find the slope of the anode curve at the op point. I used the 250v 25ma op point that was in the spec, and by eye tried to traw a tangent to the grid curve at that point. Thats all the sloping red line is, the slope of the curving grid line at that point.

Now we need to find the slope of the line, and the simple way to find the slope is find how much it moves up (ma) for a distance along (V), so I picked two arbitrary points by extending the tangent. The first point was extending down until it crosses the x axis, and I think that's about the 190v point (ish), then we just need another point, so I extended the line the other way upwards to a convenient point, in this case 500v. Does not matter where it is along the sloping line, but 500v was a good place. Then run a line across to see what current this point corresponds to, in this case 120ma.

So now we know that for a change of voltage of 500 - 190 volts, or 310v the current changed from 0 to 120ma. So the slope of the line represents a change of current with voltage. And we know from ohms law again that V / I = R. We have V and I, so R = 310 / 0.12 and this gives 2580 ohms. The 0.12 is the 120ma converted to amps.

Hope that makes some sort of sense.
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#40

Post by Nick »

I thought I'd cracked it for minute and could just use the ohms law triangle 250V/0.025A to give me the ra but it works out as 10K Ohm
Just to fill in hy that didn't work. What you have done there is calculate the static resistance, and the reason you get a different number is its going through the axis which is what you expect with resistor, the current doesn't stop flowing until there is no voltage. But we are trying to find the dynamic resistance (little r not big R) and whats different about that is that it doesn't go through the axis.

If you can imagine the slope of the tangent as you follow the grid line down, the slope decreases as the anode resistance increases, so its not like a resistor. Its a resistor whose resistance changes depending on the voltage across it.
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Dave the bass
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#41

Post by Dave the bass »

4.45AM I woke up thinking of numbers and lines, bum! I've got 'maths' baaaaad.

Thanks for your patience Nick.

Right, lets bash this down into little steps, a tangent isn't a word or 'thing' I use much so lets just make sure I've got this bit right, bearing in mind I haven't done anything like this since I left/was asked to leave school.

The operating point 250 V and 25mA came from the 'max' values on the data sheet. At that O.P. on the graph there's that lovely curve on the graph of the -32 grid bias line which I can see 'suggests' a sloping line if you mark that point and draw a line at 90 degrees to the point where that point is on the curve, so I think what you've done is this...
Image

Yes? Is that right? Hope so.

Right, next point was I just couldn't work out where you were getting the figure of 310V from, I now think its from subtracting 190V from 500V. OK, now I know where you've got the 310 figure from I can see how you've used ohms law to work out the 2583.3333 figure that is our ra, dynamic resistance, small 'r' 'cos we can't touch it physically, its a characteristic of the valve.

So, I should be able to prove that again yes? on the same slope? I've modded the gif below with my blue lines.
Image

i picked this point as its easy to read off the x and y axis where the lines cut a value of Volts and milliamps.

Using the example you did in your post using my blue lines.... I've got my 190V point and my new 350v point. So I take away 190V from 350V which = 160V. At the point where my blue line crosses your red slope (the loadline?) I've scribed a blue line across to the y axis. It cuts it at approx halfway between 50mA and 75mA so I've guessed its 62.5mA.

Using ohms law I can do 160V/0.0625A = 2560ohms. Now thats pretty close to your figure of 2583 isn't it.

I'll do another, Purple this time.
Image
Not as easy to read but it looks like the figures my purples cross are 425V and 90mA ish. So, 425V minus 190V = 235V. Therefore 235V/0.090A = 2611.111 . Again the answer is very close-ish to your 2583, it could be my dodgy scale reading.

We're onto summat, I can smell it.

Breakfast time.

Need brain food!

DTB
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#42

Post by Nick »

Yedp, thats it, I just went until the 500v so at least one part was exact (ish)

You can also pick any point for the lower value, just subtract the lower current value from the upper one to find the "distance" in ma in the same way as you are doing with the voltage.

This might help

http://en.wikipedia.org/wiki/Tangent
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#43

Post by Dave the bass »

Whahoo! Your perseverance is paying off:)

I had a thought on the way in, these are MAX parameters aren't they?

That O.P. also puts us in the bendy curvy zone doesn't it?

Could I pick a more linear, I think thats the word, O.P. and then draw a Tangent (get me!) point?

Sleepy of Dartford.

DTB
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#44

Post by Nick »

Dave the bass wrote:Whahoo! Your perseverance is paying off:)

I had a thought on the way in, these are MAX parameters aren't they?

That O.P. also puts us in the bendy curvy zone doesn't it?

Could I pick a more linear, I think thats the word, O.P. and then draw a Tangent (get me!) point?

Sleepy of Dartford.

DTB
You can pick any point in the graph, its simpler if its on a grid line, as that makes finding the tangent simpler. I just used those points as I found them in the spec sheet (not used the valve myself).
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#45

Post by Nick »

It might be hard to get a linear class A point from that valve, looks like its designed for class C use.
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